My officemate and I are currently procrastinating on research/grading (by doing some math, obviously), and he came up with this statement, which we are both convinced is false, but are having trouble coming up with a good counterexample:
Suppose $G_1$, $G_2$, $A$ are subgroups of a group $G$, $G_1\unlhd G_2$, $A\unlhd G$, and $G_2/G_1$ is abelian. Is it then true that $(A\cap G_2)/(A\cap G_1)$ is necessarily abelian?
It seems as if there should be some nice way of taking $G_1\cap A$ to be trivial and $G_2\cap A$ large enough that the quotient should be nonabelian, but writing down such a group is difficult. A counterexample involving finite groups would be especially appreciated.
Thanks in advance!
The obvious homomorphism from $A\cap G_2$ to $G_2/G_1$ has precisely $A\cap G_1$ as kernel, hence $(A\cap G_2)/(A\cap G_1)$ is isomorphic to a subgroup of $G_2/G_1$ and therefore abelian.